Discrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture


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1 Discrete Systems Mark Cannon Hilary Term 22  Lecture 4 Step response and pole locations 4 
2 Review Definition of transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k}, g k = pulse response Construct a discrete model of a continuous sampleddata system G(s) by computing the pulse response g k and transforming to get G(): { } G(s) G() = ( )Z s 42 Output response: Y () = G()U() y k = g k u k Review Analyse/design a discrete controller D(): by considering the purely discrete time system: Closed loop system tranfer function: How do the closed loop poles relate to Y () R() = G()D() + G()D() stability? performance? 43
3 Response of 2nd order system Consider the transform of a decaying exponential signal: y(t) = e at cos(bt) U(t) (U(t) = unit step) sample: y(kt ) = r k cos(kθ) U(kT ) with r = e at & θ = bt transform: Y () = 2 = ( re jθ ) + 2 ( re jθ ) ( r cos θ) ( re jθ )( re jθ ) e.g. y k is the pulse response of G(): G() = ( r cos θ) ( re jθ )( re jθ ) poles: eros: { = re jθ = re jθ { = = r cos θ 44 Response of 2nd order system 45 Responses for varying r: r < exponentially decaying envelope r = sinusoidal response with 2π/θ samples per period r > exponentially increasing envelope y k y k y k.5 r =.7 θ = π/ sample k.5.5 r =. θ = π/ sample k 5 r =.3 θ = π/ sample k
4 Response of 2nd order system 46 Responses for varying θ: θ = decaying exponential θ = π/2 2π/θ = 4 samples per period θ = π 2 samples per period y k y k y k.5 r =.7 θ = sample k.5 r =.7 θ = π/ sample k.5.5 r =.7 θ = π sample k Response of 2nd order system Some special cases: for θ =, Y () simplifies to: Y () = = exponentially decaying response r when θ = and r = : = unit step Y () = when r = : = unit pulse Y () = when θ = and < r < : samples of alternating signs 47
5 Pole positions in the plane Poles inside the unit circle are stable Im() Poles outside the unit circle are unstable Poles on the unit circle are oscillatory Real poles at < < give exponential response Higher frequency of oscillation for larger θ Re() Lower apparent damping for larer θ and r 48 Relationship with splane poles If F (s) has a pole at s = a then F () has a pole at = e at consistent with = e st What about transfer functions? { } G(s) G() = ( )Z s F(s) f(kt ) F () s s 2 s + a (kt ) kt e akt T ( ) 2 e at kt e akt T e at (s + a) 2 ( e at ) 2 a s(s + a) e akt ( e at ) ( )( e at ) b (s + a)(s + b) e akt e bkt (e at e bt ) ( e at )( e bt ) If G(s) has poles s = a i then G() has poles = e a it but the eros are unrelated a s 2 + a 2 sin akt b (s + a) 2 + b 2 e akt sin bkt sin at 2 (2 cos at ) + e at sin bt 2 2e at (cos bt ) + e 2aT 49
6 The mapping from splane to plane Locus of s = σ + jω under the mapping = e st : imaginary axis (s = jω, σ = ) unit circle ( = ) lefthalf plane (σ < ) inside of unit circle ( < ) righthalf plane (σ > ) outside of unit circle ( > ) region of splane within the Nyquist rate ( ω < π/t ) entire plane splane Im(s) plane Im() ω = π/t = e st ω = π/t Re(s) ω = ±π/t Re() 4  The mapping from splane to plane splane Im(s) = e st Im() plane s = σ + jω σ = constant Re(s) Re() = e σ e jω = e σt = constant Im(s) Im() = e st s = σ + jω ω = constant Re(s) Re() = e σ e jω arg() = ωt constant 4 
7 The mapping from splane to plane Pole locations for constant damping ratio ζ < s 2 + ζω s + ω 2 = Im(s) ζ2 ω s = ζω ± j ζ 2 ω ζω θ Re(s) cos θ = ζ ζ =.7 ζ =.5 Im(s) = e st Im() ζ =.7 ζ =.7 ζ =.5 Re(s) s = ζω + j ζ 2 ω : ζ = constant Re() ζ =.7 = e ζω T e j ζ 2 ω T 42 The mapping from splane to plane 43
8 The mapping from splane to plane ω =.5π/T ζ =.2 ω =.3π/T ζ = Second order step responses (e.g. see HLT) Design criteria based on step response: Damping ratio ζ in range.5.9 [applicationdependent] Natural frequency ω as large as possible [for fastest response] 45
9 Typical specifications for the step response: Rise time (% 9%): t r.8/ω Peak overshoot: M p e πζ/ ζ 2 Settling time (to %): t s = 4.6/(ζω ) Steady state error to unit step: e ss Phase margin: φ P M ζ 46 Typical specifications for the step response: t r, M p ζ, ω locations of dominant poles t s radius of poles: <. T /t s e ss final value theorem: e ss = lim ( )E() 46
10 Example A continuous system with transfer function G(s) = is controlled by a discrete control system with a ZOH s(s + ) The closed loop system is required to have: step response overshoot: M p < 6% step response settling time (%): t s < s steady state error to unit ramp: e ss < Check these specifications if T = s and the controller is u k =.5u k + 3(e k.88e k ) 47. (a) Find the pulse transfer function of G(s) plus the ZOH G() = ( )Z { G(s) } = s e.g. look up Z{a/s 2 (s + a)} in tables: G() = = ( ) ( ) {. } Z s 2 (s +.) ( ) (. + e. ) + ( e..e. ).484( ) ( )(.948).( ) 2 ( e. ) (b) Find the controller transfer function (using = shift operator): 48 U() E() = D() = 3 (.88 ) ( +.5 ) = 3 (.88) ( +.5)
11 2. Check the steady state error e ss when r k = unit ramp e ss = lim k e k = lim ( )E() E() R() = + D()G() R() = T ( ) so { T } e ss = lim ( ) ( ) 2 + D()G() T = lim.484( ) ( ) ( )(.948) D() = ( )D() =.96 = e ss < (as required) = lim Output y and reference r T ( )D()G() 5 Time (sec) 3. Step response: overshoot M p < 6% = ζ >.5 settling time t s < = <. / =.63 The closed loop poles are the roots of + D()G() =, i.e. (.88).484( ) + 3 ( +.5) ( )(.948) = = =.88,.5 ± j.34 But the pole at =.88 is cancelled by controller ero at =.88, and { r =.3, θ =.73 =.5 ± j.34 = re ±jθ = ζ =.56.5 Output y and input u/.5.5 all specs satisfied! Time (sec)
12 Fast sampling revisited For small T : = e st = + st + (st ) 2 /2 + + st = s T Hence the image of the unit circle under the map from to splane becomes Im( ) Im [ ( )/T ] Re( ) Re [ ( )/T ] plane loci of constant ζ & ω splane loci near = but the dominant poles lie near = so the discrete response tends to the continuous response as T 42 Summary Dependence of system pulse response on pole locations For a sampled data system with a ZOH: if s = a i is a pole of G(s), then = e a it is a pole of G() Locus of s = σ + jω under the mapping = e st : the left half plane (σ < ) maps to the unit disk ( < ) splane poles with damping ratio ζ, natural frequency ω map to plane poles with: = e ζω T arg() = ζ 2 ω T Design specifications (rise time, settling time, overshoot) imply constraints on locations of dominant poles 422
13 What you should know.... How control systems are affected by the presence of a computer in the control loop. [L] 2. How to approximate fast sampling continuous systems (filters or controllers) using discrete time approximations to continuous time derivatives. [L] 3. The effects of sample rate on a computercontrolled system. [L] 4. How to describe difference equations using transfer functions between signals represented by transforms. [L2] 5. The transfer function is the transform of the pulse response and the system output is the convolution of the pulse response and the input. [L2] 6. How to derive the discrete time model of a continuous time sampled data system using transform techniques. [L3] 7. How to compute the dynamic response of a sampled data system. [L3] 8. Properties of transforms: linearity, initial and final value theorems, multiplication, convolution and differentiation. [L3] 9. How the location of plane poles affects the step response of a second order system. [L4]. How the poles of sampled data systems map from the splane to the plane. [L4]. How to relate specifications on damping and speed of response to specifications on plane pole locations. [L4] 423
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